Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 675.
Sunday, October 2, 2011
Problem 675: Quadrilateral, 45 and 90 Degrees, Circumcircles, Intersecting Circles, Secant
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This one still doesn't have a solution? I can think of a VERY long one using: 1) Inversion with A as center and AC as radius; 2) Analytical geometry; and 3) Tons of trignometric transform. But I would love to see a shorter, synthetic solution.
ReplyDeleteLets assume midpoint of AG is J and AC is K. O,O',J are collinear points and line joining them is perpendicular to AG. If we can prove that K is collinear with O,O',J then we can prove that A,O,H are colinear using following facts.
ReplyDeleteKJ and CG are prallel and also perpendicular to AG. If CG cuts cricle O at point H, then Angle AGH is 90 deg, hence AH is diameter and A,O,H are collinear.
To prove collinearity of O,O' and K
We can drop perpendiculars from O,O'and K to ABE and ADF.
Intercept for OO' on ABE = BE/2
And for OK = BC/2sqrt (2)
Intercept for OO' on ADF = DF/2
And for OK = CD/2sqrt (2)
Since Tr.BCE and Tr.DCF are similar
By AAA similarity, BE/BC= DF/CD, hence ratio of intercepts is equal and this proves O,O' and K are collinear.